# Aptitude - Probability - Discussion

Discussion Forum : Probability - General Questions (Q.No. 4)
4.
What is the probability of getting a sum 9 from two throws of a dice?
 1 6
 1 8
 1 9
 1 12
Explanation:

In two throws of a dice, n(S) = (6 x 6) = 36.

Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}. P(E) = n(E) = 4 = 1 . n(S) 36 9

Discussion:
51 comments Page 1 of 6.

Laiba said:   2 years ago
As we know one dice has 6 sides(4 edges + 1frontside + 1backside = 6)
Possibilities of getting sum 9 are (6,3)(5,4)(4,5)(3,6) i.e 4 possibilities.
Because;
6+3 = 9
5+4 = 9
4+5 = 9
3+6 = 9.

Remember! The numbers you add should not exceed the 6 digit.

Solution for your question is @Dhaval.
4/6 * 6 * 6 = 4/216 or 1/54.
Hope,This Helps!

Qaisar wagay said:   2 years ago
Here how could you consider (5, 4) and (4, 5) two different events? Please explain.

Ankita said:   3 years ago
Why not taking the case {3,3}? Please explain me.

Nikolay said:   4 years ago
@Pooja Gm.

Dice does not have 8 surfaces it has only 6.
(1)

Surya said:   4 years ago
One dice have six faces so here we have two dices, so the formula is n2 (n Square). So 6*6 =36.

RRK said:   4 years ago
Is throwing a dice two times and throwing two dice together, same?

Arif said:   4 years ago
At first you draw sample space then you find pair which sum is 9, (1,1), (1,2),....... (5,4), (6,3),(3,6), (4,5).

Total pair 36 and sum of 9 pair is 4 them p(sum 9)= 4 ÷ 36 = 1/9.
(1)

Joshua said:   4 years ago
Because in a throw of two dices there is only 6x6, is the last the number won't Or should not exceed 6.

Swathi said:   5 years ago
Why we should not use the permutations here?like sum 9 can appear in either (6, 3) or (3, 6) , (5, 4) or (4, 5). Why we only choose the one possibility?

Pooja gm said:   5 years ago
Why we take this case only?
(3+6)
(4+5)
(5+4)
(6+3).

This also possible;
(7+2),
(1+8),
(2+7),
(8+1),